How to do summation or how to find another representation of the sum that runs over integer compositions. $$ \sum_{r_1+ \ldots + r_{L}= k} \left(-\frac{ a}{1+a} \right)^{k-L} n^{(3 r_1)} (n+3 r_{1}-1)^{(3 r_{2})}\cdot \ldots \cdot (n+\sum_{i=1}^{L-1} 3 r_i- L+1)^{(3 r_{L})} $$ where $\sum r_i = k$ runs over all integer compositions of $k$ and $L$ is a number of parts in the composition and $n^{(3 r_1)}$ is the rising factorial. $n$ is a positive integer or zero. I mean usual definition of compositions as in https://en.wikipedia.org/wiki/Composition_(combinatorics).

Realization in Mathematica may become somehow useful

Sum[(-(a/(1 + a)))^(p - r) Product[Pochhammer[n + Plus@@Table[3 k[[i]]-1, {i, 1, j - 1}],3 k[[j]]],{j, 1, r}],{r, 1, p},{k,Compositions[p-r,r]+1}]
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    $\begingroup$ According to your formula $L=k-1$. Do you mean that some of the $r_i$ are zero and $L$ is the number of nonzero ones? $\endgroup$ Aug 18, 2021 at 12:32
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    $\begingroup$ is $n^{(3)}:=n(n+1)(n+2)$ or $n(n-1)(n-2)$ ? $\endgroup$ Aug 18, 2021 at 12:44
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    $\begingroup$ $ n^{(3)}=n(n+1)(n+2)$ is right one. I mean only rising factorials. $\endgroup$ Aug 18, 2021 at 12:48
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    $\begingroup$ Your HeavisideTheta[j - 1/2] is always $1$ since your j is always a natural number $\endgroup$ Aug 18, 2021 at 15:07
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    $\begingroup$ Does this look as it should?$$\begin{array}{cccc} 6 & \frac{360 (1-a)}{a+1} & \frac{15120 (5 a^2-14 a+5)}{(a+1)^2} & \frac{5443200 (1-a) (7 a^2-30 a+7)}{(a+1)^3} \\ 24 & \frac{720 (4-3 a)}{a+1} & \frac{120960 (5 a^2-17 a+8)}{(a+1)^2} & \frac{10886400(-35 a^3+214 a^2-259 a+64)}{(a+1)^3} \\ 60 & \frac{2520 (5-3 a)}{a+1} & \frac{907200(3 a^2-12 a+7)}{(a+1)^2} & \frac{59875200(-35 a^3+243 a^2-345 a+105)}{(a+1)^3} \\ 120 & \frac{20160 (2-a)}{a+1} & \frac{1814400(5 a^2-23 a+16)}{(a+1)^2} & \frac{239500800(-35 a^3+272 a^2-443 a+160)}{(a+1)^3} \\ \end{array}$$ $\endgroup$ Aug 18, 2021 at 15:48

1 Answer 1


Let $n$ be fixed.

The sum in question can rewritten as $$S_k:=\frac{1}{(n-1)!}\sum_{L=1}^k\sum_{r_1+\dots+r_L=k} (n+3k-L)!\cdot \alpha^{k-L}\cdot f(n,k,L),$$ where $\alpha:=-\frac{a}{a+1}$ and $$f(n,k,L) := \sum_{0<s_1<\dots<s_{L-1}<k}\ \prod_{i=1}^{L-1} (n+3s_i-i).$$ (think of $s_i = r_1+\dots+r_i$)

This function for $L>1$ satisfies the recurrence: $$f(n,k,L) = \sum_{t=1}^{k-1} f(n,t,L-1)\cdot (n+3t-L+1).$$ For the generating function $$F(x,y) := \sum_{k=1}^\infty \sum_{L=1}^k f(n,k,L) x^k y^{n+3k-L}$$ it implies a linear differential equation: $$F(x,y) = \frac{xy^3}{1-xy^3}F'_y(x,y)+\frac{y^{n+2}x}{1-xy^3}$$ with a known solution.

Going back to the original sum, we apply Laplace transform to derive:

$$S_k = \frac{1}{(n-1)!\alpha^n}\ [x^k] \int_0^\infty F\big(\frac{x}{\alpha^2}, \alpha t\big) e^{-t} {\rm d}t,$$ where $[x^k]$ is the operator taking the coefficient of $x^k$.

I did not try much to simplify the result, but it gives a closed form expression for (the generating function of) $S_k$ nevertheless.

  • $\begingroup$ Thank you for this algebra, it is very impressive, but I need other thing. I start from differential equation and after solving it I obtain discret partial equation. Half of its solution I presented here. But my goal is in mathoverflow.net/questions/402164/… . $\endgroup$ Aug 20, 2021 at 15:54

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